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astra-53 [7]
3 years ago
11

Help please! This has to do with the Pythagorean theorem.

Mathematics
1 answer:
puteri [66]3 years ago
7 0

Answer:

x = 11.49

Step-by-step explanation:

Apply the pythagorean theorem to find x

Thus:

a² + b² = c²

Where,

a = x = ???

b = 8

c = 14 (hypotenuse)

Plug in the values

x² + 8² = 14²

x² + 64 = 196

x² = 196 - 64 (subtraction property of equality)

x² = 132

x = √132

x = 11.49 (approximated to nearest hundredth)

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What is the largest possible integral value in the domain of the real-valued function
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Answer:

Max Value: x = 400

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<u>Algebra I</u>

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<u>Calculus</u>

  • Antiderivatives
  • Integral Property: \int {cf(x)} \, dx = c\int {f(x)} \, dx
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Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = \frac{1}{\sqrt{800-2x} }

<u>Step 2: Identify Variables</u>

<em>Using U-Substitution, we set variables in order to integrate.</em>

u = 800-2x\\du = -2dx

<u>Step 3: Integrate</u>

  1. Define:                                                                                                            \int {f(x)} \, dx
  2. Substitute:                                                                                         \int {\frac{1}{\sqrt{800-2x} } } \, dx
  3. [Integral] Int Property:                                                                                     -\frac{1}{2} \int {\frac{-2}{\sqrt{800-2x} } } \, dx
  4. [Integral] U-Sub:                                                                                           -\frac{1}{2} \int {\frac{1}{\sqrt{u} } } \, du
  5. [Integral] Rewrite:                                                                                          -\frac{1}{2} \int {u^{-\frac{1}{2} }} \, du
  6. [Integral - Evaluate] Reverse Power Rule:                                                 -\frac{1}{2}(2\sqrt{u}) + C
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